Differential integral solution of convolution-like equation

Question

I try to find a closed form for the variable $x$ in the equality $x(t, \tau) = \int\limits_{\tau}^t \, f(t, s) x(s) ds$, without success. If you want a less general example, consider the linear multivariable example $\dot{x} = Ax + Bu$, whose solution for any $u$ is $x(t) = e^{A(t-\tau)} x_\tau + \int_{\tau}^{t} e^{A(t-s)} B u(s) ds$ and for u equal to $K x(\cdot)$ is $x(t, \tau) = e^{(A+BK)(t-\tau)} x_\tau$. The main motivation for such statement is the closed solution for differential equation $\dot{x}(t) = (\alpha(t) \, A + \beta(t) \, B) \, x(t)$.

Answer 1

It is not necessary (because slightly confusing) to include the start time $\tau$ into the solution $x(t)$ of $\dot x=Ax+Bu\,,\,\,x(\tau)=x_\tau\,.$ It is trivial that this linear inhomogenous ODE is equivalent to $$ x(t)=e^{A(t-\tau)}x_\tau+\int_\tau^te^{A(t-s)}\,B\,u(s)\,ds\quad\quad (1) $$ and that for $u(s)=Kx(s)$ we have a homogenous ODE $\dot x=(A+BK)x\,,$ with solution $x(t)=e^{(A+BK)(t-\tau)}x_\tau\,.$

It looks like you want to generalize (1) to $$ x(t)=x_0+\int_0^tf(t,s)\,x(s)\,ds\quad\quad (2) $$ (I assume $\tau=0$ for simplicity). The integral equation (2) is equivalent to the integro differential equation \begin{align} \dot x(t)&=f(t,t)\,x(t)+\int_0^t\partial_tf(t,s)\,x(s)\,ds\,\\ &=\int_0^t\Big\{f(t,t)\,f(t,s)+\partial_tf(t,s)\Big\}\,x(s)\,ds\,. \end{align} This looks more complicated than (2). However, when $\partial_tf(t,s)=A(t)f(t,s)$ we have again a homogeneous ODE $$ \dot x(t)=f(t,t)\,x(t)+A(t)\,x(t) $$ with solution $$ x(t)=x_0\exp\Big(\int_0^tf(s,s)+A(s)\,ds\Big)\,. $$ For general $f$ I cannot think of a closed form solution. Also if -as you assumed- $x_0=0$ in (2) then the only solution is the trivial one $x\equiv0\,.$